# Lagrangian of a mattress

Just started with QFT from Zee and am already confused by first equation lol. See attached picture. Does anyone actually understand this? He calls q_a the vertical displacement of particle 'a', and yet he only allows the springs to be horizontally between the particles. So, there should be no vertical displacement. Unless he envisions the mattress as oriented vertically...but that would be unnecessarily silly.

Also, why is he summing twice over the product of q's in the second term in the lagrangian? If the only springs are horizontally between the particles, and we are actually looking at horizontal instead of vertical displacement, then this second term should be,

∑(kab)(qa-qb)

I don't think this is a typo however, as the third term also contains a product.
I also attached his drawing of a mattress. Clearly in this configuration the only motion is horizontal, not vertical.

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## Answers and Replies

DrClaude
Mentor
The arrangement of the springs is in a plane, but there is no confinement to 2D. Nothing stops me from taking one of the masses and lifting it up a bit, then letting go and observing what happens.

The arrangement of the springs is in a plane, but there is no confinement to 2D. Nothing stops me from taking one of the masses and lifting it up a bit, then letting go and observing what happens.

Ok sure, but I still don't understand what the product q_a*q_b means. Why on earth would we multiply the vertical displacements of two consecutive particles? This doesn't give anything physically meaningful.

Also, I don't really understand how a 2D lattice is supposed to be motivation for a 3D field. We should have a 3D lattice with springs in every direction (x,y,z), and the particles should be free to move horizontally and vertically, not just mostly vertically. And who says a spring can't connect particles diagonally or in some other funny shape? I feel I must be missing something, as surely the foundation for a theory which makes such great predictions can't be so illogical and arbitrary..

DrClaude
Mentor
It comes from the Taylor expansion of the potential energy V:
$$V = V(0) + \sum_a \left. \frac{\partial V}{\partial q_a} \right|_0 q_a + \frac{1}{2} \sum_{a,b} \left. \frac{\partial^2 V}{\partial q_a \partial q_b} \right|_0 q_a q_b + \ldots$$
You can choose the zero of energy such that ##V(0)=0##, and since the ##q##'s are displacements, ##\left.\frac{\partial V}{\partial q_a} \right|_0 = 0## by definition of the equilibrium position (minimum energy). That gives you the first term of the potential energy in the Lagrangian. The others after that are the higher order ones.

• bhobba
DrClaude
Mentor
Also, I don't really understand how a 2D lattice is supposed to be motivation for a 3D field. We should have a 3D lattice with springs in every direction (x,y,z), and the particles should be free to move horizontally and vertically, not just mostly vertically. And who says a spring can't connect particles diagonally or in some other funny shape? I feel I must be missing something, as surely the foundation for a theory which makes such great predictions can't be so illogical and arbitrary..
You can't visualize a displacement in the fourth dimension. This is why Zee starts with a 2D lattice, so you can see the displacement along the third dimension. This is not the foundation of QFT, it is a pedagogical approach to get the students to have a sense of what excitations in a field "look" like, in order to get a better understand of the theory later on.

It comes from the Taylor expansion of the potential energy V:
$$V = V(0) + \sum_a \left. \frac{\partial V}{\partial q_a} \right|_0 q_a + \frac{1}{2} \sum_{a,b} \left. \frac{\partial^2 V}{\partial q_a \partial q_b} \right|_0 q_a q_b + \ldots$$
You can choose the zero of energy such that ##V(0)=0##, and since the ##q##'s are displacements, ##\left.\frac{\partial V}{\partial q_a} \right|_0 = 0## by definition of the equilibrium position (minimum energy). That gives you the first term of the potential energy in the Lagrangian. The others after that are the higher order ones.
I have never seen the Taylor expansion written like this, and I still don't see the reason for multiplying two vertical displacements, or even doing a Taylor expansion in the vertical direction, if there is no spring in the vertical direction in the first place..

Do you remember this form of the Taylor series from some book or have you found it on some site that you could share?

DrClaude
Mentor
I still don't see the reason for multiplying two vertical displacements, or even doing a Taylor expansion in the vertical direction, if there is no spring in the vertical direction in the first place.
Imagine that the world is 2D. How do you represent the excitation of a field? If you move one of the masses along one of those two dimensions, it would correspond to some strange warping of space. The excitation takes place locally in this 2D world, so you can represent it as some motion in a third dimension outside this 2D world. Think for instance of an electromagnetic wave. The electric and magnetic feilds are often represented as oscillations in planes perpendicular to the propagation of the EM wave, but these planes are not along an actual dimension of space. If you have an EM wave moving perfectly horizontally, you don't have an electric field oscillating up and down; the wave itself is infinitely small along the vertical direction. It is the same here: you need an extra dimension to visualize what is a local oscillation.

Do you remember this form of the Taylor series from some book or have you found it on some site that you could share?
You can check it out on Wikipedia.

Imagine that the world is 2D. How do you represent the excitation of a field? If you move one of the masses along one of those two dimensions, it would correspond to some strange warping of space. The excitation takes place locally in this 2D world, so you can represent it as some motion in a third dimension outside this 2D world. Think for instance of an electromagnetic wave. The electric and magnetic feilds are often represented as oscillations in planes perpendicular to the propagation of the EM wave, but these planes are not along an actual dimension of space. If you have an EM wave moving perfectly horizontally, you don't have an electric field oscillating up and down; the wave itself is infinitely small along the vertical direction. It is the same here: you need an extra dimension to visualize what is a local oscillation.

You can check it out on Wikipedia.
Ok thanks so much for your patience! I think I get it now:D